2007 AMC 10A Problems/Problem 9

Revision as of 12:02, 23 July 2017 by Progamexd (talk | contribs) (Solution 2)

Problem

Real numbers $a$ and $b$ satisfy the equations $3^{a} = 81^{b + 2}$ and $125^{b} = 5^{a - 3}$. What is $ab$?

$\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60$

Solution 1

\[81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8\]

And

\[125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b\]

Substitution gives $4b+8 - 3 = 3b \Longrightarrow b = -5$, and solving for $a$ yields $-12$. Thus $ab = 60\ \mathrm{(E)}$.

Solution 2

Simplify equation $1$ which is $3^a=81^b+2$, to $3^a=3^4(b+2)$, which equals $3^a=3^4b+8$.

And

Simplify equation $2$ which is $125^b=5^a-3$, to $5^3(b)=5^a-3$, which equals $5^3b=5^a-3$.

Now, eliminate the bases from the simplified equations $1$ and $2$ to arrive at $a=4b+8$ and $3b=a-3$. Rewrite equation $2$ so that it is in terms of $a$. That would be $a=3b+3$.

Since both equations are equal to $a$, and $a$ and $b$ are the same number for both problems, set the equations equal to each other. $4b+8=3b+3$ $b=-5$

Now plug $b$, which is $(-5)$ back into one of the two earlier equations. $4(-5)+8=a$ $-20+8=a$ $a=-12$

$(-12)(-5)=60$

Therefore the correct answer is E

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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